inlining higher-order-functions?

I've just discovered the {-# INLINE #-} pragma, but it's not doing as much for me as I had hoped. My example is complicated, so let me present a simpler analogy. Suppose I defined compose :: (b -> c) -> (a -> b) -> (a -> c) compose f g = \x -> f (g x) I can easily persuade GHC to inline 'compose'. But when 'compose' is applied to known arguments, I wish f and g to be inlined in the body of 'compose'. Is there a pragma that will do the trick? (I attempted to put an INLINE pragma in a where clause, but GHC was not amused.) Norman

| My example is complicated, so let me present a simpler analogy. | Suppose I defined | | compose :: (b -> c) -> (a -> b) -> (a -> c) | compose f g = \x -> f (g x) | | I can easily persuade GHC to inline 'compose'. | But when 'compose' is applied to known arguments, I wish | f and g to be inlined in the body of 'compose'. | Is there a pragma that will do the trick? | (I attempted to put an INLINE pragma in a where clause, | but GHC was not amused.) You can put inline pragmas on f and g, thus frob = ... {-# INLINE frob #-} burk = ... {-# INLINE burk #-} wibble = compose from burk Now compose will be inlined (assuming it too has an INLINE pragma), and then frob, burk. Simon

Hello Norman, Friday, December 22, 2006, 8:23:57 AM, you wrote:
compose :: (b -> c) -> (a -> b) -> (a -> c) compose f g = \x -> f (g x)
ghc 6.6 added 'inline' function, see user docs. although only SPJ knows whether it can be used here: compose f g = inline (\x -> f (g x)) -- Best regards, Bulat mailto:Bulat.Ziganshin@gmail.com
participants (3)
-
Bulat Ziganshin
-
nr@eecs.harvard.edu
-
Simon Peyton-Jones